In evaluating contributions to the two-point function in say $\phi^3$ theory to:
$$\langle 0|\phi(x)\phi(y)e^{-i\int d^4z\frac{\lambda}{3!}\phi^3(z)}|0\rangle,$$
at $\mathcal{O}(\lambda^2)$, one of the possible contractions is the usual tadpole diagram. However, the literature often says that diagrams which can be rendered disconnected with a single cut do not contribute to this matrix element (Collins Renormalization pg. 41, Peskin & Schroeder pg. 219).
My question: Does this mean that tadpoles don't contribute to the self energy since one can separate the bubble from the source with a cut? That doesn't sound right to me, but maybe I could keep them but also include a counterterm
$$\langle 0|\phi(x)\phi(y)e^{i\int d^4z\left(\frac{\lambda}{3!}\phi^3(z)+c\phi\right)}|0\rangle.$$
One could generate an $\mathcal{O}(\lambda^2)$ contribution via the mixed term $$-\int d^4z_1 d^4z_2\frac{\lambda^2 c}{3!}\phi^3(z_1)\phi(z_2)$$ which would generate something like a tadpole counterterm contribution to the two-point function. I suppose my confusion can be summed up as follows:
-
If I don't disregard tadpoles in the two-point function, must I include the counterterm $c\phi$ in the interaction Lagrangian?
-
If so, does the counterterm remove the entire diagram anyway, or just the divergent part (assuming the contribution is finite + divergent)?
Best Answer
Yes, in general the self-energy $\Sigma=G_0^{-1}-G_c^{-1}$ may contain tadpoles$^1$ even though they are not 1PI.
However, if one imposes the renormalization condition $\langle \phi \rangle_{J=0}=0$, then one may show that the self-energy only contains 1PI diagrams, and hence no tadpoles, cf. my Phys.SE answer here.
--
$^1$ NB: Note that a self-loop diagram is not necessarily a tadpole diagram, cf. Wikipedia.